Question

If specific heat capacity of water 4 200 J/kg 0C, heat of fusion of ice 336 000 J/kg, so specific heat capacity of ice is

Answer 1

Explanation:

For water, m

1

=200,c

1

=4200,t

1

=50

o

C

For Ice, m

2

=40,c

2

=336000

Heat required to melt ice is mass×Latent heat of fusion=40×10

−3

×336×10

3

=13440J.

Let final temperature be T,

For water, m=200×10

−3

,c=4200,t=50

So energy supplied by water is 4200×200×10

−3

×(50−T)

After ice is melted to water, m=40×10

−3

,c=4200,t=0

So energy absorbed is 4200×40×10

−3

×T

Balancing energy,

4200×200×10

−3

×(50−T)=13440+4200×40×10

−3

×T

Solving above, T=28.33

o

C